How to redraw circuits in this problem to see series/paralle

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In summary, the problem is asking to find the equivalent resistance, R_T, in the given circuit diagram for the cases where terminals a and b are open circuited and short circuited. The solution provided in the textbook involves combining the top two and bottom two resistors in series, which results in two parallel combinations. In the short circuited case, the 40 Ω and 60 Ω resistors are in series, as are the 90 Ω and 10 Ω resistors. By redrawing the circuit, it becomes clear that c and d are connected in series in the open circuit case, and in parallel in the short circuited case. The problem statement is important to include for clarity and to
  • #1
LongApple
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Homework Statement



See pic for problem statement and diagram (we are dealing with the diagram on the right)
http://i.imgur.com/Uv7Ikma.png
I'll rewrite the problem statement here if that helps: Find R_T in the diagram for the cases where ab is open circuited and short circuited.

And its solution in the textbook: The first part of is part a) the second part b)
http://i.imgur.com/yYLGoFE.png

Why is it that after combining the top two resistors and combining the bottom two resistors in the picture that the two resulting series combinations are parallel to each other?

Homework Equations


None, this is conceptual

The Attempt at a Solution


part a)
I drew what I think combining the top two and bottom two resistors would look like here in part a for the open circuit case:
http://i.imgur.com/Uv7Ikma.png
In the pic with the shadow of the hand below, I draw more pictures and I dont' see why the circuit has anything parallel
http://i.imgur.com/azqgWwD.png
The picture drawn on the right labeled "parallel" was just for me to visually compare the left to see where the "two series combinations are in parallel" in the part a) solutions

but I don't see how it leads to the statement in the solutiosn "The two series combinations are in parallel" for part a) which is the first solution in the solution imgur linkIn part b) the short circuited case, I have also drawn what I believe to be equivalent circuits in the short circuited case but don't understand why the 40/60 combo and 90/10 combo mentioned in the solutions are in series
http://i.imgur.com/h4X5Dcn.png

What are some smarter ways to draw the circuits to better see why certain combinations are in parallel or series?
 
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  • #2
What are some smarter ways to draw the circuits to better see why certain combinations are in parallel or series?
To avoid errors like your last picture (*), you can flip the 90 Ω and 10 Ω resistors from left to right with the line ab as axis. The it becomes clear that 40 and 90 are in series, as are 60 and 10 (case ab open).



(*) you draw a circle around a non-connection and move it to the top left as if that were short-circuited too (that would yiled RT=0 !)

By te way, PF guidelines don't appreciate this way of posting with a lot of links. How long will the imgur pictures be available ?

I see in another thread you started that you might even be in favour of expiring links to avoid copyright problems. But PF aims to build up a long-lasting 'database' of re-usable threads. One more reason to use picture uploads (preferable) and picture links only as a supplement, not as a substitute.
 

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  • #3
redraw.JPG
https://www.physicsforums.com/attachments/77023

You can walk from c to a and then from a to d (blue path). If nothing else is connected to a, the 40 Ω and 90 Ω resistors are connected in ?
You can also reach from c to d along the green path. If noting is connected to b the 60 Ω and the 10 Ω resistors are connected in ?
So you can redraw the circuit as in the picture on the right.
What happens if a and b are short-circuited?
 
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  • #4
ehild said:
https://www.physicsforums.com/attachments/77023

You can walk from a to c and then from c to b (blue path). If nothing else is connected to c, the 40 Ω and 60 Ω resistors are connected in ?
You can also reach from a to b along the green path. If noting is connected to d the 90 Ω and the 10 Ω resistors are connected in ?
So you can redraw the circuit as in the picture on the right.
What happens if c and d are short-circuited?

I just want to confirm that the intersection X of blue and green and in circuit diagrams in general does not mean there is a physical intersection. Am I correct?
So like the blue could come out of the page and the green could go under the page?
 
  • #5
LongApple said:
I just want to confirm that the intersection X of blue and green and in circuit diagrams in general does not mean there is a physical intersection. Am I correct?
So like the blue could come out of the page and the green could go under the page?
Yes

@ehild: the exercise wants RT, not Rab
 
  • #6
LongApple said:
I just want to confirm that the intersection X of blue and green and in circuit diagrams in general does not mean there is a physical intersection. Am I correct?
So like the blue could come out of the page and the green could go under the page?
There is no physical intersection in the middle of the drawing. If it was, it would be drawn by a circle, as it would be a node.
Yes, the lines can go one above and the other below the paper.
 
  • #7
BvU said:
Yes

@ehild: the exercise wants RT, not Rab
Sorry, I misread ...I attached a corrected picture. I hope, it i right now.
 
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  • #8
ehild said:
https://www.physicsforums.com/attachments/77023

You can walk from a to c and then from c to b (blue path). If nothing else is connected to c, the 40 Ω and 60 Ω resistors are connected in ?
You can also reach from a to b along the green path. If noting is connected to d the 90 Ω and the 10 Ω resistors are connected in ?
So you can redraw the circuit as in the picture on the right.
What happens if c and d are short-circuited?

I just want to make sure I understand the problem statement

http://i.imgur.com/Uv7Ikma.png

In part b) when they say short circuited they mean connecting terminals a and b, correct? Or is it c and d?
 
  • #9
Sorry, I misread the problem. I corrected my original post and picture. The problem means a and b short-circuited. You have to find how the resistors are connected with respect to the terminals c and d.

It is very important that you write the problem statement in your post. People can misread and forget what the question was, when the question is not in front of their eyes.
 
  • #10
LongApple said:
I just want to make sure I understand the problem statement

http://i.imgur.com/Uv7Ikma.pngUv7Ikma[1].png

In part b) when they say short circuited they mean connecting terminals a and b, correct? Or is it c and d?
 

FAQ: How to redraw circuits in this problem to see series/paralle

How do I determine if the circuit is in series or parallel?

To determine if the circuit is in series or parallel, you can look at the connection of the components. In a series circuit, the components are connected one after the other, creating a single path for current to flow. In a parallel circuit, the components are connected in branches, creating multiple paths for current to flow.

Can I redraw a circuit to see it in both series and parallel?

Yes, you can redraw a circuit to see it in both series and parallel. To see it in series, you can draw the components in a single line. To see it in parallel, you can draw the components in branches, with each branch connecting to a common point on either side.

How do I redraw a circuit to see series/parallel if the components are not all the same?

If the components in the circuit are not all the same, you can still redraw the circuit to see it in series/parallel. However, you may need to use different symbols for the components, such as a resistor with a different resistance value. The overall connections of the components should still follow the rules of series and parallel circuits.

Can I use Kirchhoff's laws to analyze a circuit in both series and parallel?

Yes, Kirchhoff's laws can be used to analyze a circuit in both series and parallel. However, you may need to apply the laws separately to each section of the circuit, depending on how it is redrawn.

Is it necessary to redraw a circuit to see it in series/parallel?

No, it is not always necessary to redraw a circuit to see it in series/parallel. However, redrawing the circuit can make it easier to analyze and understand the connections and flow of current. It can also help in simplifying the circuit for further calculations or modifications.

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